<script>
$(document).ready(function(){
var id = 1;
var name = "Abu";
$.post("update.php", {id: id, name: name}, function(data)
{
});
});
</script>
update.php
<?php
$id = $_POST['id'];
$name = $_POST['name'];
echo $id.'<br />';
echo $name.'<br />';
?>
From the above code, I want to use jquery pass the id and name from js to the update.php to update the record. However, I get the following error: Undefined index .... What am I doing wrong?
hii michael you can set the ajax post request like this
$.ajax({
type: 'POST',
url: 'abc.php',
data: {box:id,keys:name},
success: function(msg) {
alert(msg);
}
});
then in php file you can check with isset if value is set or not
You can also test other approaches (since you still don't show more details from your element inspector)with following things:
1.Name attributes in POST data with serialized string:
var data = "name=nameString";
$.post("file", data, function(data) {
}, "json");
2.Or use map:
var data = {
name : "name"
};
$.post("file", data, function(data) {
}, "json");
To handle this variable in PHP , keep the same use:
$name = $_POST['name'];
Related
I need help on how to post and retrieve multiple variables using Ajax Post. I actually needed to retrieve the posted variables for SQL query. See below the Ajax Code where i needed to include variable names selschool, selprogram, selsession to the post
<script>
$("#session").change(function()
{
$("#loding2").show();
var id=$(this).val();
var dataString = 'id='+ id;
var selschool=document.getElementById("selectedschool").val();
var selprogram=document.getElementById("selectedprogram").val();
var selsession=document.getElementById("selectedsession").val();
$("#semester").find('option').remove();
$("#class").find('option').remove();
document.getElementById("selectedclass").value= " ";
document.getElementById("selectedsemester").value= " ";
$.ajax
({
type: "POST",
url: "get_class.php",
data: dataString,
cache: false,
success: function(html)
{
$("#loding2").hide();
$("#class").html(html);
}
});
});
</script>
Also see below PHP script where i wanted to use the posted variable for the query;
<?php
include('dbconfig.php');
if($_POST['id'])
{
$id=$_POST['id'];
// Todo: I actually needed something like where session SELECT * FROM class where session_id=$id and program_id="selprogram" and school_id="selschool"
$stmt = $DB_con->prepare("SELECT * FROM class where session_id=$id ");
$stmt->execute(array(':id' => $id));
?><option selected="selected">Select Class :</option>
<?php while($row=$stmt->fetch(PDO::FETCH_ASSOC))
{
?>
<option value="<?php echo $row['class_id']; ?>"><?php echo $row['class_name']; ?></option>
<?php
}
}
?>
Let me explain the solution
consider the ajax call to demo.php
$.ajax({
url: 'demo.php',
type: 'post',
data: {
'name': 'abc',
'phone': '1234567899'
}, //data is in json form
success: function(res) {
console.log(JSON.parse(res)); //parsing because we will pass the data from demo.php in encoded form you will get it.
}
});
now in demo.php you will access data as $_POST['name'] and $_POST['phone']. lets pass the same to ajax call. will store it in array and will pass it.
<?php
$Arr = [];
$Arr[0] = $_POST['name'];
$Arr[1] = $_POST['phone'];
echo json_encode($Arr);
?>
like this, we can pass data to ajax and can pass the data from PHP file to request.
Hope that you got the result. Thank you.
Hello im using a login script with ajax and i want to callback and display my data
email and usernam to stock it in local storage.
i can get the data in json but i want to display this data in console.log
this is my codes
send_ajax.js
$(document).ready(function(e) {
$("#contactSubmitButton").click(function(){
var email = $("#contactEmailField").val();
var password = $("#contactNameField").val();
$.ajax({
type: "POST",
url: "http://hubafrica.co/webservices/get_user.php",
data: "email="+email+"&password="+password,
dataType: "json",
cache: false,
beforeSend: function(){ $("#contactSubmitButton").val('Chargement...');},
success: function(data) {
alert(data);
if(data)
{
iterateJson(data);
var url="http://hubafrica.co/webservices/get_user.php";
$.get(url,function(data){
// loop through the members here
$.each(json.data,function(i,dat){
console.log(dat.email);
window.localStorage.setItem("id", dat.id);
});
});
//window.location.href = "user_dashboard.html";
}else{
$("#formSuccessMessageWrap").fadeIn('slow');
$("#contactSubmitButton").val('se connecter');
}
},
error: function (xhr, ajaxOptions, thrownError) {
}
});
});
});
script.php
<?php
header("Content-Type:application/json");
header('Access-Control-Allow-Origin: *');
include("../config/config.php");
$account=array();
if (isset($_POST["email"])){
$email = htmlspecialchars($_POST["email"]);
$pass = htmlspecialchars($_POST["password"]);
$sql = mysql_query('SELECT * FROM `b2b_user` where email="'.$email.'" and password="'.$pass.'"');
$num = mysql_num_rows($sql);
if($num > 0){
$row = mysql_fetch_assoc($sql);
$account['id'] = $row['id'];
$account['email'] = $row['email'];
echo '{"members":'.json_encode($account).'}';
}
}
?>
be for send response from backend you need to formate your data in json. Once you get in response you need to parseJSON()and the you can menuplate.
change here:use data.members to access data:
in your php construct like this:do not append string.
$account1= array('members'=>$account);
echo json_encode($account1);
script:
$.each(data.members,function(i,dat){
console.log(dat.email);
window.localStorage.setItem("id", dat.id);
});
You can either use
alert(JSON.stringify(data));
or
Console.log(JSON.stringify(data));
Following line of code turns an object/array in to a JSON text and stores that JSON text in a string.
JSON.stringify('/array variable here/')
Hope it helps.
I have an array that i pass from javascript to php and in php page i am trying to put it in session to be used in the third page. The code is as below
JavaScript:
var table_row = [];
table_row[0] = [123,123,123];
table_row[1] = [124,124,124];
table_row[2] = [125,125,125];
var jsonString = JSON.stringify(table_row);
$.ajax({
type: "POST",
url: "test1.php",
dataType: "json",
data: {myJSArray: jsonString},
success: function(data) {
alert("It is Successfull");
}
});
test1.php
<?php
session_start();
$check1 = $_POST['myJSArray'];
$_SESSION['array']= $check1;
echo $check1;
?>
test2.php
<?php
session_start();
$test = $_SESSION['array'];
echo $test;
?>
on submit i call the function in javascript and the form takes me to test2.php. It is giving error on test2.php page Notice: Undefined index: array in C:\xampp\htdocs\test\test2.php on line 13
Any suggestions please do let me know.
You don't need to stringify yourself, jquery does it for you, if you stringify it, jQuery will believe you want a string instead
var table_row = [];
table_row[0] = [123,123,123];
table_row[1] = [124,124,124];
table_row[2] = [125,125,125];
$.ajax({
type: "POST",
url: "test1.php",
dataType: "json",
data: {myJSArray: table_row},
success: function(data) {
alert("It is Successfull");
}
});
However, on the php side, you still need to decode it as it is always a string when you get it from $_POST. use json_decode to do it.
$check1 = json_decode($_POST['myJSArray']);
look at your test2.php
<?php
session_start();
$test = $_SESSION['array'];
echo $test;
?>
if it's only the code in the file then the error you got C:\xampp\htdocs\test\test2.php on line 13 is mindless, because there is not line 13,
but if you have something about the code you show us, may there be something echoed before?
because session has to be started before any output,
otherwise I've tested whole script and works fine...
To check if session really started (otherwise $_SESSION will not work), try this:
if(session_id())
{
echo "Good, started";
}
else
{
echo "Magic! strangeness";
}
if problem not found in test2.php you can check test1.php echo $_SESSION['array'] after saving it, and in your javascript callback function alert data param itself,
I'm sure you can catch the problem by this way.
i got it to work, the code is below
Javascript file: in page index.php
Either you can call this function and pass parameter or use code directly
var table_row = []; //globally declared array
var table_row[0]=["123","123","123"];
var table_row[1]=["124","124","124"];
var table_row[2]=["125","125","125"];
function ajaxCode(){
var jsonArray = JSON.stringify(table_row)
$.ajax
({
url: "test1.php",
type: "POST",
dataType: 'json',
data: {source1 : jsonArray},
cache: false,
success: function (data)
{
alert("it is successfull")
}
});
}
Page: test1.php
session_start();
unset($_SESSION['array']);
$check1 = $_POST['source1'];
$_SESSION['array']= $check1;
echo json_encode(check1);
Page: test2.php //final page where i wanted value from session
if(session_id())
{
echo "Session started<br>";
$test = $_SESSION['array'];
echo "The session is".$test;
}
else
{
echo "Did not get session";
}
?>
In index page i have a form that is submitted and on submission it calls the ajax function.
Thank you for the help, really appreciate it.
Below is my ajax call code. I want to send one data in .php file via ajax call and want to get two values from .php file. This two values I want to set in different 'input' tags whose id are 'course_name' and 'course_credit'.
Here my ajax call return correct value(real value from DB table) of 'course_name' input tag.
But 'MY PROBLEM IS' the value of input tag whose id is 'course_credit' shows 'success'. How can I get the correct value(real value from DB table) of id 'course_credit' ?
I have a 'select' tag which id is 'c_select'
HTML:
<input type="text" name="course_name" id="course_name" value=""/>
<input type="text" name="course_credit" id="course_credit" value=""/>
AJAX :
$('#c_select').change(function(){
$.ajax({
type:'post',
url:'get_course_info_db.php',
data: 'c_id='+ $(this).val(),
success: function(reply_data1,reply_data2){
$('#course_name').val(reply_data1);
$('#course_credit').val(reply_data2);
}
});
});
get_course_info_db.php
<?php
include('db_connection.php');
$c_id = $_POST['c_id'];
$result = mysql_query("SELECT * FROM course WHERE c_id = '$c_id'");
$all_course_data = mysql_fetch_array($result);
$c_name = $all_course_data['c_name'];
$c_credit = $all_course_data['c_credit'];
echo $c_name,$c_credit;
exit();
?>
AJAX code:-
$('#c_select').change(function(){
$.ajax({
type:'post',
url:'get_course_info_db.php',
data: 'c_id='+ $(this).val(),
success: function(value){
var data = value.split(",");
$('#course_name').val(data[0]);
$('#course_credit').val(data[1]);
}
});
});
PHP code:-
<?php
include('db_connection.php');
$c_id = $_POST['c_id'];
$result = mysql_query("SELECT * FROM course WHERE c_id = '$c_id'");
$all_course_data = mysql_fetch_array($result);
$c_name = $all_course_data['c_name'];
$c_credit = $all_course_data['c_credit'];
echo $c_name.",".$c_credit;
exit();
?>
The success callback is Function( PlainObject data, String textStatus, jqXHR jqXHR ); http://api.jquery.com/jQuery.ajax/
php:
$data = array(
'name' => $c_name,
'credit' => $c_credit,
);
echo json_encode($data);
javascript:
success: function(data) {
var result = $.parseJSON(data);
$('#course_name').val(result.name);
$('#course_credit').val(result.credit);
}
success: function(reply_data1,reply_data2){
$('#course_name').val(reply_data1);
$('#course_credit').val(reply_data2);
}
second arguement is the status of http request, you have to encode the answer, i suggest you JSON
in your php
$c_credit = $all_course_data['c_credit'];
echo json_encode(array('name' => $c_name,'credit' => $c_credit));
exit();
and in your javascript
success: function(response,status){
var datas = JSON.parse(response);
$('#course_name').val(datas.name);
$('#course_credit').val(data.credit);
}
this is not tested, but this is the way to do it
I'd suggest using JSON to encode the data you fetch from the database.
Try changing your ajax call as follows:
$('#c_select').change(function(){
$.ajax({
type:'post',
url:'get_course_info_db.php',
data: 'c_id='+ $(this).val(),
dataType: 'json', // jQuery will expect JSON and decode it for you
success: function(reply_data){
$('#course_name').val(reply_data['c_name']);
$('#course_credit').val(reply_data['c_credit']);
}
});
});
And your PHP as follows:
include('db_connection.php');
// Escape your input to prevent SQL injection!
$c_id = mysql_real_escape_string($_POST['c_id']);
$result = mysql_query("SELECT * FROM course WHERE c_id = '$c_id'");
$all_course_data = mysql_fetch_array($result);
echo json_encode($all_course_data);
exit();
I haven't tested this but I imagine it'd work for you.
i am trying to get a row of data from database using ajax in codeigniter.
Here is the javascript function-
$(function(){
$("button[name='program_view_details']").click(function(e){
e.preventDefault();
var program_id=$(this).attr('id');
$.ajax({
url: "<?php echo base_url();?>program_management/get_program_data",
type: "POST",
dataType: "html",
data: "program_id="+program_id,
success: function(row)
{
alert(row.program_name);
}
});
});
I am not sure if the datatype and post is correct or not.
Here is my controller function-
public function get_program_data( ){
$program_id = $this->input->post('program_id');
$this->load->model('program_management_model');
$data['programs']= $this->program_management_model->get_program_specific($program_id);
echo $data;
}
Here is the model-
function get_program_specific($program_id){
$query=$this->db->query("SELECT * FROM programs WHERE program_id='".$program_id."'");
return $query->result();
}
I am searching the way of returning the row from controller to javascript. But the alert() is showing "undefined" in the success. Please anyone tell me the whole way through. Thanks in advance.
$data['programs']= $this->program_management_model->get_program_specific($program_id);
The $data which you are echoing in the controller is basically an array[] and programs is an array which is present in $data. Either echo the $data in controller using the
foreach(){}
or echo the $query array in your model. That will do the trick.And in ajax success call just append the data to the element in which you want to show the result.
Change names as per your page.
in your script:
$.ajax({
url: '<?php echo base_url();?>managealerts_edit/editalerts',
type: "POST",
data: {'id': edit_id},
cache: false,
dataType: "json",
success: function(row){
//alert(row.sub);
$('#edit').show();
$('#sub').val(row.sub);
$('#mess').val(row.mess);
}
});
in your model:
$query = $this->db->query("SELECT fld_id, fld_course_id,fld_sub,fld_mess from tbl_alerts where fld_id='".$det."' ");
if ($query->num_rows() > 0)
{
$row = $query->row_array();
$data=array("sub" => $row['fld_sub'], "mess" => $row['fld_mess']);
echo json_encode($data);
}
in your controller:
$det = $this->input->post('id');
//$alertsres['tbl_alerts'] = $this->managealerts_m->select_editalerts($det);
$this->managealerts_m->select_editalerts($det);`
in model
function get_program_specific($program_id){
$temp=array();
$query=$this->db->query("SELECT * FROM programs WHERE program_id='".$program_id."'");
$temp= $query->row_array();
echo $temp['program_name'];
}
in controller change the line
$data['programs']= $this->program_management_model->get_program_specific($program_id);
with
$this->program_management_model->get_program_specific($program_id);
and finally in javascript
alert(row);
please let me know if you face any problem.